# Integral of a differential form

• kiuhnm
In summary, the conversation discusses a proof that if a smooth differential ##n-1##-form ##\omega## on ##\mathbb{R}^n## is ##0## outside of a ball of radius ##R##, then the integral of its differential ##d\omega## over all of ##\mathbb{R}^n## is also ##0##. This is shown by using the fact that ##\omega## is also ##0## on the surface of the ball, and that the partial derivatives of the functions involved are also ##0## outside of the ball. The proof ultimately involves using the equation $$\oint_{\partial K} \omega = \int_K d\omega$$ and the

## Homework Statement

Suppose that a smooth differential ##n-1##-form ##\omega## on ##\mathbb{R}^n## is ##0## outside of a ball of radius ##R##. Show that $$\int_{\mathbb{R}^n} d\omega = 0.$$

## Homework Equations

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$$\oint_{\partial K} \omega = \int_K d\omega$$

## The Attempt at a Solution

If ##\omega## is ##0## outside of the ball, by continuity, it must be ##0## on the surface of the ball as well. We know that $$\omega = \sum_{i=1}^n f^i(x^1,\ldots,x^n) dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^n$$ for some functions ##f^i:\mathbb{R}^n\to\mathbb{R}##, so
$$d\omega = \sum_{i=1}^n (-1)^{i-1} \frac{\partial f^i}{\partial x^i} dx^1\wedge\cdots\wedge dx^n.$$
All those partial derivatives are ##0## outside of the ball, so $$\int_{\mathbb{R}^n} d\omega = \int_B d\omega = \oint_{\partial B} \omega = 0,$$ where ##B## is the ball.